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Katyanochek1 [597]
3 years ago
6

hydraulic lift is to be used to lift a 2100-kg weight by putting a weight of 25 kg on a piston with a diameter of 10 cm. Determi

ne the diameter of the piston on which the weight is to be placed.
Physics
1 answer:
trasher [3.6K]3 years ago
3 0

Answer:

840 cm

Explanation:

Note: A hydraulic press operate based on pascal's principle.

From pascal's principle

W₁/d₁ = W₂/d₂...................... Equation 1

Where W₁ and W₂ are the first and second weight, and d₁ and d₂ are the  first and second diameter of the piston.

make d₁ the subject of the equation

d₁ = W₁×d₂/W₂................ Equation 2

Given: W₁ = 2100 kg, W₂ = 25 kg, d₂ = 10 cm = 0.1 m.

Substitute these values into equation 2

d₁ = 2100(0.1)/25

d₁ = 8.4 m

d₁ = 840 cm

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Find the acceleration a body whose velocity increases from 11m/s to 33m/s in 10 seconds
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Answer:

1.1 m/(s)^2

Explanation:

u=11 m/s

v=33 m/s

t=10s

v=u+at

=> 33=22+(a)(10)

=> 33-22=10a

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2 years ago
On a snowy day, max (mass = 15 kg) pulls his little sister maya in a sled (combined mass = 20 kg) through the slippery snow. max
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Work done by a given force is given by

W = F.d

here on sled two forces will do work

1. Applied force by Max

2. Frictional force due to ground

Now by force diagram of sled we can see the angle of force and displacement

work done by Max = W_1 = Fdcos\theta

W_1 = 12*5cos15

W_1 = 57.96 J

Now similarly work done by frictional force

W_2 = Fdcos\theta

W_2 = 4*5cos180

W_2= -20 J

Now total work done on sled

W_{net}= W_1 + W2

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7 0
3 years ago
A DJ starts up her phonograph player. The turntable accelerates uniformly from rest, and takes t1 = 11.9 seconds to get up to it
Degger [83]

Answer:

a)\omega_1=8.168\,rad.s^{-1}

b)n_1=7.735 \,rev

c)\alpha_1 =0.6864\,rad.s^{-2}

d)\alpha_2=4.1454\,rad.s^{-2}

e)t_2=1.061\,s

Explanation:

Given that:

  • initial speed of turntable, N_0=0\,rpm\Rightarrow \omega_0=0\,rad.s^{-1}
  • full speed of rotation, N_1=78 \,rpm\Rightarrow \omega_1=\frac{78\times 2\pi}{60}=8.168\,rad.s^{-1}
  • time taken to reach full speed from rest, t_1=11.9\,s
  • final speed after the change,  N_2=120\,rpm\Rightarrow \omega_2=\frac{120\times 2\pi}{60}=12.5664\,rad.s^{-1}
  • no. of revolutions made to reach the new final speed,  n_2=11\,rev

(a)

∵ 1 rev = 2π radians

∴ angular speed ω:

\omega=\frac{2\pi.N}{60}\, rad.s^{-1}

where N = angular speed in rpm.

putting the respective values from case 1 we've

\omega_1=\frac{2\pi\times 78}{60}\, rad.s^{-1}

\omega_1=8.168\,rad.s^{-1}

(c)

using the equation of motion:

\omega_1=\omega_0+\alpha . t_1

here α is the angular acceleration

78=0+\alpha_1\times 11.9

\alpha_1 = \frac{8.168 }{11.9}

\alpha_1 =0.6864\,rad.s^{-2}

(b)

using the equation of motion:

\omega_1\,^2=\omega_0\,^2+2.\alpha_1 .n_1

8.168^2=0^2+2\times 0.6864\times n_1

n_1=48.6003\,rad

n_1=\frac{48.6003}{2\pi}

n_1=7.735\, rev

(d)

using equation of motion:

\omega_2\,^2=\omega_1\,^2+2.\alpha_2 .n_2

12.5664^2=8.168^2+2\alpha_2\times 11

\alpha_2=4.1454\,rad.s^{-2}

(e)

using the equation of motion:

\omega_2=\omega_1+\alpha_2 . t_2

12.5664=8.168+4.1454\times t_2

t_2=1.061\,s

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